Archive for June, 2013

"What [the National Weather Service] has found, year after year, is that the levels of carbon pollution in our atmosphere have increased dramatically. That science, accumulated and reviewed over decades, tells us that our planet is changing in ways that will have profound impacts on all of humankind." Speech delivered by President Obama at Georgetown University, Washington D.C. on Tuesday 25th June 2013.

“What [the National Weather Service] has found, year after year, is that the levels of carbon pollution in our atmosphere have increased dramatically. That science, accumulated and reviewed over decades, tells us that our planet is changing in ways that will have profound impacts on all of humankind.”
Speech delivered by President Obama at Georgetown University, Washington D.C. on Tuesday 25th June 2013.

In the week (week 26, 2013 – Ed) that President Obama announced his bold plan for the US to reduce atmospheric carbon pollution and put the brakes on global warming, what better than a blog post that examines the thermochemical relationship between heat production and CO2 emissions in hydrocarbon fuels.

But before we begin…

cg

The camping gas cartridge pictured above contains a mixture of butane (C4H10) and propane (C3H8). Combustion of this mixture generates heat, along with the combustion products carbon dioxide and water.

Now here’s a question for you. If instead of the mixture, the cartridge contained only butane, would you expect combustion to produce more or less carbon dioxide in relation to the amount of heat generated?

Base your reasoning solely on the information given above. Jot down your answer, and we will come back to it later.

– – – –

Thermochemistry is the branch of thermodynamics which studies thermal processes associated with chemical reactions and other forms of chemical change. From its beginnings in the 18th century, thermochemistry played an important role in overcoming the misunderstandings about heat which arose from its confusion with the concept of temperature.

Laplace and Lavoisier's ice calorimeter of 1780. James Clerk Maxwell wrote of it: "Nothing can be more perfect than the theory and design of this apparatus. It is worthy of Laplace and Lavoisier, and in their hands it furnished good results."

Laplace and Lavoisier’s ice calorimeter of 1780. James Clerk Maxwell wrote of it: “Nothing can be more perfect than the theory and design of this apparatus. It is worthy of Laplace and Lavoisier, and in their hands it furnished good results.”

Through the development and use of the calorimeter, two important early discoveries were made. In France in 1780, those men of genius – Antoine Lavoisier and Pierre-Simon Laplace – discovered that the heat gained or lost in a chemical reaction is equal in magnitude and opposite in sign for the reverse reaction. And in Russia in 1840, the Swiss-born Germain Hess discovered that if a given set of reactants is transformed into a given set of products by more than one reaction path, the total amount of heat gained or lost is the same for each path.

The second of these two discoveries, now called Hess’s law, has for decades been peculiarly subject to factual distortion in science education. Hess measured heat changes at constant pressure (qp), but because of the later-discovered identity qp = ΔH, Hess’s law is commonly stated in terms of enthalpy change. Students are thus misled into thinking that Hess discovered an enthalpy summation relation, when in fact he had been dead for 25 years before the concept of enthalpy was first defined by Gibbs in 1875. This kind of timeline falsification, although motivated by an understandable desire to keep things simple in textbooks and internet resources, does the proper study of thermodynamics a disservice in my view.

– – – –

The development in the first half of the 20th century of the concept of the covalent bond led directly to a new class of thermochemical data called bond enthalpies. Tables of these averaged quantities allow the estimation of enthalpy of reaction using the rule:

ΔHrxn = ΣΔH(bonds broken) – ΣΔH(bonds formed)

Care needs to be taken with average values, since there can be considerable variations in certain circumstances and you need eyes in the back of your head for things like phase transitions and unusual oxidation states, but for applications involving uncomplicated single-phase reactions, bond enthalpies can be a very useful resource.

Here’s an example using an analytical technique of my own. Consider the combustion of alkanes in the vapor phase, which can be represented by the general equation:

CnH2n+2(g) + ½(3n+1)O2(g) → nCO2(g) + (n+1)H2O(g)

Say you are interested in the relationship between the heat generated by the reaction (ΔHrxn) and the number of moles (n) of carbon dioxide produced.

The procedure is as follows. First, expressions in n are found for bonds broken and formed:

bonds broken

total

avg. bond enthalpy

C–H

2n+2

a

O=O

½(3n+1)

b

C–C

n-1

c

bonds formed  

 

C=O

2n

d

O–H 2n+2

e


The enthalpy of reaction ΔHrxn can then be expressed as the sum of the bond enthalpies of the reactants, minus the sum of the bond enthalpies of the products, all bond enthalpies being regarded as positive quantities.

ΔHrxn =  [(2n+2)a + ½(3n+1)b + (n-1)c] – [(2n)d + (2n+2)e]    …(1)

The incremental value of ΔHrxn for an integer increase in n can be found by differentiation:

d(ΔHrxn)/dn = [2a + (3/2)b + c] – [2d + 2e]    …(2)

It is evident from inspection of (1) that (2) represents the limiting value of ΔHrxn per mole of CO2 produced as n becomes very large. This fixes the far end of the alkane scale; the other end is found by setting n=1 in (1):

ΔHrxn(n=1) = [4a + 2b] – [2d + 4e]    …(3)

By inserting the values of the average bond enthalpies into expressions (3) and (2), the range of ΔHrxn per mole of CO2 produced by the combustion of alkanes is determined. The absolute values may not be very accurate, but the trend as n increases will have sufficient accuracy to indicate which alkanes generate the most energy per mole of CO2 produced.

Note: The analytical technique described above is original work by the author of CarnotCycle. If you make use of it for your own purposes, please include a reference to this blog/blogpost. Many thanks.

– – – –

At a time when atmospheric CO2 levels have passed the 400 ppm mark for the first time since the Pliocene period, it seems an appropriate moment to take a look at the principal hydrocarbon fuels and see which of them generate most energy in relation to carbon dioxide produced.

Energy generated (kJ)

Reaction enthalpy

per gram fuel per mole O2

per mole CO2

Natural gas CH4 + 2O2=CO2 + 2H2O

810

51.6 405

810

Ethanol C2H5OH + 3O2 = 2CO2 + 3H2O

1257

27.3 419

628

Petroleum 2(-CH2-) +3O2 = 2CO2 + 2H2O

1220

43.6 407

610

Coal 4(-CH-) + 5O2 = 4CO2 + 2H2O

2046

39.3 409

512

Cellulose (-CHOH-) + O2 = CO2 + H2O

447

14.9 447

447


According to these figures widely quoted on the internet, natural gas gives the best quotient among the fossil fuels, delivering 42% more energy per mole of CO
2 produced than coal, which explains the “cleaner alternative” phrase favored by energy ministers when asked about all the sudden interest in shale gas.

In terms of transport fuels, there is little difference on this analysis between ethanol and petroleum. The value of bioethanol supplementation of gasoline would appear to be linked more to sustainability and supply security than energy/emission quotient, although for fair comparison one should take biomass origin and CO2 emissions during industrial processing into the overall reckoning.

Cellulose gives the lowest quotient, although again it must be acknowledged that the carbon it contains came from atmospheric CO2 in the first place. The only observation I would make about the popular “CO2-neutral” appellation is that while living plants generally take up CO2 at average atmospheric concentrations, power stations burning biomass pump out CO2 at vastly higher concentrations. Under continual operation, a steady-state gradient with a high local CO2 concentration is likely to be established. If there is water nearby (power stations are often placed next to large water masses for cooling purposes), CO2 will readily dissolve in it. So in overall terms, CO2-neutral could actually mean accelerated atmosphere-to-ocean transfer, which in view of the resultant ocean acidification may not be entirely neutral in its environmental impact.

– – – –

One last point to cover is the underlying reason for the differences in energy/emission quotient shown in the final column of the above table. The figures in the penultimate column for energy generated per mole of oxygen consumed show remarkably little variation between fuels, which indicates that the explanation of the difference lies in the other element in the CO2 molecule – carbon.

The covalently bonded carbon atom can vary in its oxidation state depending on the atoms other than carbon to which it is attached. Bonding to hydrogen decreases the oxidation state, while bonding to oxygen increases it. And this affects the amount of energy released from a hydrocarbon fuel on combustion – the lower the average oxidation number of the carbon, the higher the enthalpy of combustion, and vice versa. The table below demonstrates the correlation nicely.

kJ per mole of CO2

Avg. oxidation number of carbon in fuel

Natural gas CH4 + 2O2 = CO2+2H2O

810

–4

Ethanol C2H5OH + 3O2 = 2CO2 + 3H2O

628

–2

Petroleum 2(-CH2-) +3O2 = 2CO2 + 2H2O

610

–2

Coal 4(-CH-) + 5O2 = 4CO2 + 2H2O

512

–1

Cellulose (-CHOH-) + O2 = CO2 + H2O

447

0


We have reached the end, so let’s go back to that question posed at the beginning –

“Would you expect butane to produce more or less carbon dioxide in relation to the amount of heat generated than a butane-propane mixture?”

The answer is more carbon dioxide.

The average oxidation number of carbon in butane is –10/4, while that of propane is –8/3. Butane alone has a less negative (i.e. higher) average carbon oxidation number than any butane-propane mixture, and so should produce more carbon dioxide in relation to the amount of heat generated. Enthalpy of combustion determinations show this prediction to be correct.

– – – –

 

Advertisements
photo credit: theguardian.com

photo credit: theguardian.com

From the search term phrases that show up on this blog’s stats, CarnotCycle detects that a significant segment of visitors are studying foundation level thermodynamics  at colleges and universities around the world. So what better than a post that tackles that favorite test topic – exact and inexact differentials.

When I was an undergraduate, back in the time of Noah, we were first taught the visual approach to these things. Later we dispensed with diagrams and got our answers purely through the operations of calculus, but either approach is equally instructive. CarnotCycle herewith presents them both.

– – – –

The visual approach

Ok, let’s start off down the visual track by contemplating the following pair of pressure-volume diagrams:

visual track

The points A and B have identical coordinates on both diagrams, with A and B respectively representing the initial and final states of a closed PVT system, such as an ideal gas, whose state can be defined in terms of any two of the three variables p,v,t, since the third will be determined by the equation of state. In this case, the pressure P and the volume V have been selected as the independent variables.

The paths taken between A and B are visibly different. In path I on the left, the volume first increases at constant pressure (ΔV>0, external work done), then the pressure decreases at constant volume (ΔV=0, no work done). In path II on the right, these operations are performed in reverse order.

Now the PV diagram has the property that the area under the curve represents the work done during the transition from initial state to final state. This is represented by the shaded area in the diagrams; it is evident that the work done in passing from A to B is greater for path I than it is for path II.

The conclusion can thus be drawn that unless the path is specified, the work done in the process A→B cannot be determined.

Now let’s apply the first law of thermodynamics to the process A→B. For an appreciable (i.e. measurable) change

exact09

where q is the heat supplied to the system and w is the work done by the system. The internal energy U is a state function – this is an axiom of thermodynamics – and for a closed system it means that ΔU has the same value for all paths between state A and state B. Since the above equation does not specify a path, and we know that under these circumstances w is indeterminate, q must also be indeterminate.

For a PVT system undergoing a reversible process, we can write the differential equation which expresses the first law as follows:

exact10

The same considerations applying to appreciable differences also apply to infinitesimals on integration. For the process A→B, the values of the integrals ∫ABdqrev and ∫ABdwrev are path-dependent.

Now the mathematics of multivariate functions teaches us that if the differential of a function Y, where Y = Y(X1, X2,…, Xn), can be set equal to a differential expression ΣCidXi where Xi are independent variables and the coefficients Ci are functions of the Xi, then dY is said to be an exact differential.

exact07

exact08

Importantly for thermodynamics, dY then also has the characteristic that the value of the integral ∫ABdY = ΔY is independent of the path followed from state A to state B. As we have seen, dU comes into this category.

If a differential does not have this characteristic, i.e. if the value of the integral is path-dependent, then it is said to be an inexact differential. And as we have seen, dqrev and dwrev come into this category.

– – – –

The math approach

Right then, let’s see how we can set about supplying the answer as to why dqrev is inexact by taking the purely mathematical route. We can write the differential equation which expresses the first law for a PVT system undergoing a reversible process in the following way:

exact01

Actually, when you look at this equation it does seem reasonable to think that dqrev is exact, since both the variables on the right are exact. But appearances can be deceptive.

Now since P,V and T are the thermodynamic variables which determine the state of a PVT system, and U is a state function, we can consider U a function of any two of them – the third will be determined by the equation of state. So let’s choose V and T (makes the calculus a bit easier) and write the partial differential equation:

exact02

Substituting for dU in the first law equation gives:

exact03

At this point, it is convenient to recall the property of exact differentials that if Y and its derivatives are continuous, then for any pair of independent variables X1 and X2 there is the mathematical requirement that

exact04

We can use this to test the exactness of the property relation we have just derived for dqrev. Differentiating M wrt T and N wrt V gives:

exact05

exact06

Since we know that U is a function of T and V, it follows that (∂2U/∂T∂V) and (∂2U/∂V∂T) must be equal. (∂M/∂T)V and (∂N/∂V)T are therefore not equal, and since (∂P/∂T)V is generally non-zero (i.e. not a horizontal line on a PT plot), we must conclude that the test fails to establish exactness, and that dqrev – and therefore dwrev – are inexact differentials.